$g(x)=\begin{cases} \text{ln}(x)&\text{for }0<x\leq2 \\\\ x^2\text{ln}(2)&\text{for }x>2 \end{cases}$ Find $\lim_{x\to 2}g(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\text{ln}(2)$ (Choice B) B $4$ (Choice C) C $4\cdot\text{ln}(2)$ (Choice D) D The limit doesn't exist.
Explanation: $x=2$ is on the boundary between the pieces of our piecewise function. In order to find $\lim_{x\to 2}g(x)$, we need to find the one-sided limits. Let's find the limit as $x$ approaches $2$ from the left. We will use the fact that $g(x)=\text{ln}(x)$ for $x$ -values smaller than $2$. $\begin{aligned} &\phantom{=}\lim_{x\to 2^-}g(x) \\\\ &=\lim_{x\to 2^-}\text{ln}(x) \\\\ &=\text{ln}(2)&\gray{\text{Direct substitution}} \end{aligned}$ Let's find the limit as $x$ approaches $2$ from the right. We will use the fact that $g(x)=x^2\text{ln}(2)$ for $x$ -values greater than $2$. $\begin{aligned} &\phantom{=}\lim_{x\to 2^+}g(x) \\\\ &=\lim_{x\to 2^+}x^2\text{ln}(2) \\\\ &=2^2\cdot\text{ln}(2)&\gray{\text{Direct substitution}} \\\\ &=4\cdot\text{ln}(2) \end{aligned}$ $\text{ln}(2)\neq 4\cdot\text{ln}(2)$ so the one-sided limits aren't equal. This means that the two-sided limit $\lim_{x\to 2}g(x)$ doesn't exist.